Triangulated Categories [Lecture notes]

57 Pages·2014·0.527 MB·English

by  Pawel Sosna

TRIANGULATED CATEGORIES, SUMMER SEMESTER 2012
P. SOSNA
Contents
1. Triangulated categories and functors 2
2. A first example: The homotopy category 8
3. Localization and the derived category 12
4. Derived functors 26
5. t-structures 31
6. Bondal’s theorem 38
7. Some remarks about homological mirror symmetry 43
7.1. Some notions from algebraic geometry 43
7.2. Symplectic geometry and homological mirror symmetry 44
8. Fourier–Mukai functors 45
8.1. Direct image 45
8.2. Inverse image 46
8.3. Tensor product 46
8.4. Derived dual 47
8.5. The construction and Orlov’s result 47
9. Spherical twists 49
Appendix: Stability conditions 55
References 57
Triangulated categories are a class of categories which appear in many areas of mathematics.
The following graphical representation is based on [1].
(tensor) tr. cat.
(cid:51)(cid:51) (cid:55)(cid:55) (cid:79)(cid:79) (cid:104)(cid:104) (cid:107)(cid:107)
alg. geom. repr. th. st. hom. th. motiv. th. noncomm. top.
(1) Algebraic geometry Here we could take Db(X) if X is a smooth and projective
variety or the category of perfect complexes or...
(2) Stable homotopy theory The stable homotopy category of topological spectra is a
triangulated category. The tensor product is the smash product.
1
2 P.SOSNA
(3) (Modular) representation theory Derived category of modules over an algebra. Or
we could consider a finite group G, a field k with char(k) > 0 and take the stable
module category of finitely generated kG-modules. Objects are k-representations of G
and morphisms are kG-linear map modulo those factoring through a projective.
(4) Motivic theory Voevodsky’s derived category of geometric motives.
(5) Noncommutative topology G reasonable topological group; can consider the G-
equivariant Kasparov category of separable G-C∗-algebras.
The goal of the lecture is to give a gentle introduction to some basic notions in the theory
of triangulated categories. In particular, we will consider the examples of the homotopy and
derived category of an abelian category, derived functors and t-structures.
We will need the following notions from category theory. Recall that, roughly speaking, an
additive category is a category such that the Hom-spaces have the structure of abelian groups
and the compositions are bilinear, there exists a 0-object and finite direct sums exist as well.
An additive functor is supposed to be compatible with these additional structures. Given a
field k, we can consider k-linear additive categories, meaning that we require the Hom-spaces
to be k-vector spaces and all compositions to be k-bilinear. Similarly, one defines the notion of
a k-linear functor.
1. Triangulated categories and functors
Definition. Let D be an additive category. The structure of a triangulated category on D
(cid:47)(cid:47)
is given by an additive autoequivalence T = [1]: D D, the shift functor, and a class of
distinguished triangles
(cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
A B C T(A) = A[1]
satisfying the following axioms.
TR1 (i) Any triangle of the form
id (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
A A 0 A[1]
is distinguished.
(ii) Any triangle isomorphic to a distinguished triangle is itself distinguished (a mor-
phism of triangles is a collection of vertical maps such that everything commutes).
(cid:47)(cid:47)
(iii) Any morphism f: A B can be completed to a distinguished triangle
f (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
A B C A[1].
TR2 The triangle
f (cid:47)(cid:47) g (cid:47)(cid:47) h (cid:47)(cid:47)
A B C A[1]
is distinguished if and only if
g (cid:47)(cid:47) h (cid:47)(cid:47) −f[1](cid:47)(cid:47)
B C A[1] B[1].
TRIANGULATED CATEGORIES, SUMMER SEMESTER 2012 3
TR3 Suppose there exists a commutative diagram of distinguished triangles with vertical
arrows α and β
(cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
A B C A[1]
α β γ α[1]
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
A(cid:48) (cid:47)(cid:47) B(cid:48) (cid:47)(cid:47) C(cid:48) (cid:47)(cid:47) A(cid:48)[1],
then there exists a γ making this diagram a morphism of triangles.
f (cid:47)(cid:47) g (cid:47)(cid:47)
TR4 For each pair of morphisms A B C there is a commutative diagram
A f (cid:47)(cid:47) B h (cid:47)(cid:47) C(cid:48) (cid:47)(cid:47) A[1]
= g =
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
A g◦f (cid:47)(cid:47) C (cid:47)(cid:47) B(cid:48) (cid:47)(cid:47) A[1]
f[1]
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
A(cid:48) = (cid:47)(cid:47) A(cid:48) (cid:47)(cid:47) B[1]
(cid:15)(cid:15) (cid:15)(cid:15)
B[1] h[1] (cid:47)(cid:47) C(cid:48)[1]
where the first two rows and the two central columns are distinguished triangles.
Remark 1.1. A category satisfying TR1-TR3 is called pre-triangulated. The last axiom TR4 is
usually called the octahedral axiom since it can be represented in the form of an octahedron. It
is relatively rarely used. There are no easy examples of pre-triangulated categories which are
not triangulated.
Remark 1.2. Consider two composable maps f,g as in TR4 and denote the object C from TR1
(iii)byC(f). Ifonethinksofdistinguishedtrianglesasageneralizationofshortexactsequences,
(cid:14)
then we get C(f) = B/A, C(g) = C/B and C(g◦f) = C/A. Then TR4 says that C/A B/A (cid:39)
(cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
C/B, since it asserts the existence of a triangle C(f) C(g◦f) C(g) C(f)[1] .
In the following we will derive some easy consequences of the definition.
Proposition 1.3. Let
f (cid:47)(cid:47) g (cid:47)(cid:47) h (cid:47)(cid:47)
A B C A[1]
be a distinguished triangle. Then g◦f = 0.
Proof. Consider the commutative diagram
id (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
A A 0 A[1]
id f =
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
f (cid:47)(cid:47) g (cid:47)(cid:47) (cid:47)(cid:47)
A B C A[1],
4 P.SOSNA
where the first row is a triangle by TR1 (i). We are then done by TR3. (cid:3)
f (cid:47)(cid:47) g (cid:47)(cid:47) (cid:47)(cid:47)
Proposition 1.4. Let A B C A[1] be a distinguished triangle in D. Then,
for any object A , the sequences
0
(cid:47)(cid:47) (cid:47)(cid:47)
Hom(A ,A) Hom(A ,B) Hom(A ,C)
0 0 0
and
(cid:47)(cid:47) (cid:47)(cid:47)
Hom(C,A ) Hom(B,A ) Hom(A,A )
0 0 0
are exact.
(cid:47)(cid:47)
Proof. Let h: A B be a morphism such that g◦h = 0. Apply TR1 and TR3 to
0
id (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
A A 0 A [1]
0 0 0
h
(cid:15)(cid:15)
f (cid:47)(cid:47) g (cid:47)(cid:47) (cid:47)(cid:47)
A B C A[1]
(cid:47)(cid:47)
in order to get a morphism m: A A with f ◦m = h. This proves the exactness of the first
0
sequence (im ⊂ ker follows from the previous proposition) and the second is similar. (cid:3)
Remark 1.5. We can use TR2 to apply the proposition to the rotated triangles and hence we
get long exact sequences.
Proposition 1.6. Let
(cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
A B C A[1]
α β γ α[1]
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
A(cid:48) (cid:47)(cid:47) B(cid:48) (cid:47)(cid:47) C(cid:48) (cid:47)(cid:47) A(cid:48)[1]
be a morphism of triangles. If two out of the three vertical arrows are isomorphisms, then so
is the third.
Proof. Without loss of generality assume that α and β are isomorphisms. Apply Hom(C(cid:48),−)
to both triangles to get
Hom(C(cid:48),A) (cid:47)(cid:47) Hom(C(cid:48),B) (cid:47)(cid:47) Hom(C(cid:48),C) (cid:47)(cid:47) Hom(C(cid:48),A[1]) (cid:47)(cid:47) Hom(C(cid:48),B[1])
α∗ β∗ γ∗ α[1]∗ β[1]∗
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
Hom(C(cid:48),A(cid:48)) (cid:47)(cid:47) Hom(C(cid:48),B(cid:48)) (cid:47)(cid:47) Hom(C(cid:48),C(cid:48)) (cid:47)(cid:47) Hom(C(cid:48),A(cid:48)[1]) (cid:47)(cid:47) Hom(C(cid:48),B(cid:48)[1]),
a commutative diagram with exact rows. By assumption, α , β , α[1] and β[1] are isomor-
∗ ∗ ∗ ∗
phisms, so γ is also one by the 5-lemma. Thus, there exists a map δ: C(cid:48) (cid:47)(cid:47)C such that
∗
γ ◦δ = id . Similarly one can use Hom(C,−) to conclude that there is also a left inverse.
C(cid:48)
Then they have to be equal and γ is an isomorphism as claimed. (cid:3)
Corollary 1.7. The distinguished triangle in TR1 (iii) is unique up to isomorphism. (cid:3)
TRIANGULATED CATEGORIES, SUMMER SEMESTER 2012 5
(cid:47)(cid:47)
Remark 1.8. For a given map f: A B one usually calls C in TR1 (iii) a cone of f. We
have just seen that a cone is unique up to isomorphism. But this isomorphism is not unique,
because of the non-uniqueness of the map in TR3. This seems like a minor issue but has in fact
as a consequence that certain natural constructions do not work in the context of triangulated
categories.
f (cid:47)(cid:47) g (cid:47)(cid:47) h (cid:47)(cid:47)
Proposition 1.9. Let A B C A[1] be a distinguished triangle. Then f is
an isomorphism if and only if C (cid:39) 0.
Proof. Assume f is an isomorphism and let f−1 be the inverse. Then we have the diagram
id (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
B B 0 B[1]
f−1 id
(cid:15)(cid:15) (cid:15)(cid:15)
f (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
A B C A[1].
By TR3 and Proposition 1.6 C (cid:39) 0.
If C (cid:39) 0, use the two rows in the above triangle to get f−1. (cid:3)
Lemma1.10. Let A f (cid:47)(cid:47) B g (cid:47)(cid:47) C h (cid:47)(cid:47) A[1] and A(cid:48) f(cid:48) (cid:47)(cid:47) B(cid:48) g(cid:48) (cid:47)(cid:47) C(cid:48) h(cid:48) (cid:47)(cid:47) A(cid:48)[1] bedis-
tinguished triangles. Then their direct sum is also a triangle.
Proof. Consider the map f⊕f(cid:48): A⊕A(cid:48) (cid:47)(cid:47)B⊕B(cid:48). By TR1 this can be completed to a triangle
∆ := A⊕A(cid:48) (cid:47)(cid:47)B⊕B(cid:48) (cid:47)(cid:47)P. Projectingtothefactorswegettwomorphismsoftrianglesfrom∆
to the given ones, so, in particular, we get maps P (cid:47)(cid:47)C and P (cid:47)(cid:47)C(cid:48). Thus we get a map from
∆ to the direct sum of the given triangles and Proposition 1.6 gives that the two triangles are
isomorphic(tobemoreprecise,wehavetousethatHom(C⊕C(cid:48),−) (cid:39) Hom(C,−)⊕Hom(C(cid:48),−),
that direct sums of exact sequences is exact; we then get Hom(C⊕C(cid:48),−) (cid:39) Hom(P,−), hence
P (cid:39) C ⊕C(cid:48) by Yoneda; we only need that Hom(A ,−) resp. Hom(−,A ) gives short exact
0 0
sequences). (cid:3)
f (cid:47)(cid:47) g (cid:47)(cid:47) h (cid:47)(cid:47)
Proposition 1.11. Let A B C A[1] be a distinguished triangle and assume
that h = 0. Then B (cid:39) A⊕C.
(cid:47)(cid:47) id (cid:47)(cid:47) (cid:47)(cid:47)
Proof. We know that 0 C C 0 is a triangle by TR1 (i) and TR2. There is
id (cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
also A A 0 A[1] . Taking their direct sum gives a triangle by the lemma.
Now complete the diagram
(cid:47)(cid:47) (cid:47)(cid:47) 0 (cid:47)(cid:47)
A A⊕C C A[1]
id id id
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
(cid:47)(cid:47) (cid:47)(cid:47) 0 (cid:47)(cid:47)
A B C A[1]
to a morphism of triangles, which have to be isomorphic. (cid:3)
6 P.SOSNA
Proposition 1.12. Consider the following diagram
f (cid:47)(cid:47) g (cid:47)(cid:47) h (cid:47)(cid:47)
A B C A[1]
β
(cid:15)(cid:15)
A(cid:48) f(cid:48) (cid:47)(cid:47) B(cid:48) g(cid:48) (cid:47)(cid:47) C(cid:48) h(cid:48) (cid:47)(cid:47) A(cid:48)[1].
If g(cid:48)◦β ◦f = 0, then there are maps α: A (cid:47)(cid:47)A(cid:48) and β: C (cid:47)(cid:47)C(cid:48) completing the diagram to a
morphism of triangles.
Proof. Apply Hom(A,−) to the lower triangle to get
... (cid:47)(cid:47) Hom(A,A(cid:48)) (cid:47)(cid:47) Hom(A,B(cid:48)) (cid:47)(cid:47) Hom(A,C(cid:48)) (cid:47)(cid:47) ...
By assumption, the morphism βf ∈ Hom(A,B(cid:48)) goes to zero in Hom(A,C(cid:48)), hence we get a
morphism α ∈ Hom(A,A(cid:48)) such that f(cid:48)α = βf. Then we can use TR3 to get γ. (cid:3)
Remark 1.13. Note that if Hom(A,C(cid:48)[−1]) = 0, then α and γ are unique. The first statement
being obvious, we prove the second. To do this, apply Hom(−,C(cid:48)) to the upper triangle to get
... (cid:47)(cid:47) Hom(A[1],C(cid:48)) (cid:47)(cid:47) Hom(C,C(cid:48)) (cid:47)(cid:47) Hom(B,C(cid:48)) (cid:47)(cid:47) ...
Then g(cid:48)◦β ∈ Hom(B,C(cid:48)) goes to g(cid:48)◦β◦f = 0 ∈ Hom(A,C(cid:48)), hence we get γ, which has to be
unique by assumption.
Definition. Let F: D (cid:47)(cid:47)D(cid:48) be an additive functor. Then F is called exact if
(1) There exists an isomorphism of functors F ◦T (cid:39) T ◦F.
D D(cid:48)
(cid:47)(cid:47) (cid:47)(cid:47) (cid:47)(cid:47)
(2) Any distinguished triangle A B C A[1] in D is mapped to a distin-
guished triangle F(A) (cid:47)(cid:47) F(B) (cid:47)(cid:47) F(C) (cid:47)(cid:47) F(A)[1] in D(cid:48), where we identify
F(A(cid:48)[1]) with F(A(cid:48))[1] via the functor isomorphism from (i).
Definition. A triangulated subcategory D(cid:48) of D is a full additive subcategory admitting the
structure of a triangulated category such that the inclusion functor D(cid:48) ⊂ D is exact and every
object isomorphic to an object of D(cid:48) is in D(cid:48).
Remark 1.14. The above definition is equivalent to the following one. The category D(cid:48) is
invariant under shift and if in a triangle two out of three objects are in D(cid:48), then so is the third.
Also note that we could equally well define a triangulated subcategory without the assump-
tion that it is closed under isomorphisms.
Example 1.15. Let F: D (cid:47)(cid:47)D(cid:48) be an exact functor. Then ker(F) := {A ∈ D | F(A) (cid:39) 0} is
a triangulated subcategory of D. Furthermore, if A⊕B ∈ ker(F), then A and B are in ker(F).
The second assertion being trivial, we prove the first one. Clearly, ker(F) is invariant under
shift. If we take a triangle in D and two of the three objects have the property that their image
is isomorphic to 0, the same holds for the third.
TRIANGULATED CATEGORIES, SUMMER SEMESTER 2012 7
(cid:47)(cid:47)
Definition. Let F: A B be a functor between arbitrary categories. We say that a functor
(cid:47)(cid:47)
H: B A is right adjoint to F if there exist functorial isomorphisms
Hom (F(A),B) (cid:39) Hom (A,H(B))
B A
(cid:47)(cid:47)
for all A ∈ A and B ∈ B. A functor G: B A is left adjoint to F if there exist functorial
isomorphisms
Hom (B,F(A)) (cid:39) Hom (G(B),A)
B A
for all A ∈ A and B ∈ B.
Note that given an A ∈ A, we can put B = F(A) and then the first equation gives
Hom (F(A),F(A)) (cid:39) Hom (A,HF(A)),
B A
(cid:47)(cid:47)
so we get a natural transformation id HF. Similarly, we also get a natural transformation
A
(cid:47)(cid:47)
FH id .
B
Proposition 1.16. Let F: D (cid:47)(cid:47)D(cid:48) be an exact functor between triangulated categories. If a
left (or right) adjoint functor exists, then it is exact.
Proof. We will only consider the case of a right adjoint functor H. Let us first check that
it commutes with the respective shifts. Since F is exact, we have a functorial isomorphism
F ◦T (cid:39) T(cid:48) ◦F and similarly for the inverse of the shift. This yields the following functorial
D
isomorphisms
Hom(A,H(T(cid:48)(B))) = Hom(F(A),T(cid:48)(B)) (cid:39) Hom(T(cid:48)−1F(A),B)
= Hom(FT−1(A),B) (cid:39) Hom(T−1(A),H(B))
= Hom(A,TH(B)).
By the Yoneda lemma we get an isomorphism HT(cid:48) (cid:39) TH.
Now we have to show that H maps an exact triangle in D(cid:48) to an exact triangle in D.
Let A(cid:48) (cid:47)(cid:47) B(cid:48) (cid:47)(cid:47) C(cid:48) (cid:47)(cid:47) A(cid:48)[1] be a triangle in D(cid:48) and complete the induced morphism
H(A(cid:48)) (cid:47)(cid:47)H(B(cid:48)) to a triangle in D:
H(A(cid:48)) (cid:47)(cid:47) H(B(cid:48)) (cid:47)(cid:47) C (cid:47)(cid:47) H(A(cid:48))[1],
0
where we tacitly use the isomorphism HT(cid:48) (cid:39) TH. Using the the assumption that F is exact
and the adjunction morphisms FH(A(cid:48)) (cid:47)(cid:47)A(cid:48) and FH(B(cid:48)) (cid:47)(cid:47)B(cid:48), one obtains a morphism of
distinguished triangles
FH(A(cid:48)) (cid:47)(cid:47) FH(B(cid:48)) (cid:47)(cid:47) F(C ) (cid:47)(cid:47) FH(A(cid:48))[1]
0
ξ
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
A(cid:48) (cid:47)(cid:47) B(cid:48) (cid:47)(cid:47) C(cid:48) (cid:47)(cid:47) A(cid:48)[1],
where the dotted line exists by TR3.
8 P.SOSNA
(cid:47)(cid:47)
Applying H to the diagram and using the adjunction h: id H ◦F, yields
H(A(cid:48)) (cid:47)(cid:47) H(B(cid:48)) (cid:47)(cid:47) C (cid:47)(cid:47) H(A(cid:48))[1]
0
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) hC0 (cid:15)(cid:15)
HFH(A(cid:48)) (cid:47)(cid:47) HFH(B(cid:48)) (cid:47)(cid:47) HF(C ) (cid:47)(cid:47) HFH(A(cid:48))[1]
0
H(ξ)
(cid:35)(cid:35) (cid:15)(cid:15) (cid:35)(cid:35) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
H(A(cid:48)) (cid:47)(cid:47) H(B(cid:48)) (cid:47)(cid:47) H(C(cid:48)) (cid:47)(cid:47) H(A(cid:48))[1].
It is a fact from category theory that the curved arrows are identity morphisms. Now, for any
A we consider the sequence
0
Hom(A ,H(B(cid:48))) (cid:47)(cid:47) Hom(A ,H(C(cid:48))) (cid:47)(cid:47) Hom(A ,H(A(cid:48))[1])
0 0 0
and note that it is exact due to adjunction and the exactness of F (this sequence is iso-
morphic to the one we get if we apply Hom(F(A ),−) to our original exact triangle). Then
0
Hom(A ,C ) (cid:39) Hom(A ,H(C(cid:48))) for all A and hence H(ξ) ◦ h : C (cid:47)(cid:47)H(C(cid:48)) is an iso-
0 0 0 0 C0 0
morphism. Thus, H(A(cid:48)) (cid:47)(cid:47) H(B(cid:48)) (cid:47)(cid:47) H(C(cid:48)) (cid:47)(cid:47) H(A(cid:48))[1] is isomorphic to the distin-
guished triangle H(A(cid:48)) (cid:47)(cid:47) H(B(cid:48)) (cid:47)(cid:47) C (cid:47)(cid:47) H(A(cid:48))[1] and hence is distinguished. (cid:3)
0
Remark 1.17. We will see later (Remark 3.24) that a similar statement does not hold in the
abeliansetting. Roughly,oneusesthatthetensorproductfunctorisadjointtotheHom-functor
(on the category of abelian groups, say) and the former can be exact while the latter is only
half-exact.
2. A first example: The homotopy category
Definition. Let A be an additive category. A complex in A is a diagram of objects and
morphisms in A
... (cid:47)(cid:47) Ai−1 di−1 (cid:47)(cid:47) Ai di (cid:47)(cid:47)(cid:47)(cid:47) Ai+1 di+1 (cid:47)(cid:47) ...
such that dk+1◦dk = 0 for all k ∈ Z. One often writes a complex as A•.
A complex is bounded if Ak = 0 for |k| (cid:29) 0. It is bounded below if Ak = 0 for k (cid:28) 0 and it
is bounded above if Ak = 0 for k (cid:29) 0.
AmorphismbetweentwocomplexesA• andB• consistsofacollectionofmapsfi: Ak (cid:47)(cid:47)Bk
for all k ∈ Z such that dk fk = fk+1dk for all k ∈ Z. The category of all complexes in A will
B• A•
be denoted by Kom(A), its full subcategory of bounded complexes by Komb(A), the category
of bounded below complexes by Kom+(A) and the category of bounded above complexes by
Kom−(A).
Remark 2.1. It is straightforward to check that Kom(A) is an additive category.
Remark 2.2. Note that complexes naturally arise in, for example, topology. If X is a CW-
complex, we can consider singular chains. If X is a differentiable manifold, one can consider
the de Rham complex.
TRIANGULATED CATEGORIES, SUMMER SEMESTER 2012 9
Definition. Let A• be a complex and let n be an integer. We define the complex A•[n] by
setting A•[n]k = Ak+n and dk = (−1)ndk+n. For a morphism f: A• (cid:47)(cid:47)B• one defines
A•[n] A•
f[n]: A•[n] (cid:47)(cid:47)B•[n] by setting f[n]k = fn+k.
Remark 2.3. One checks easily that [n], the n-th shift functor, defines an autoequivalence of
Kom(A) with inverse [−n].
Definition. A morphism f: A• (cid:47)(cid:47)B• is called homotopic to zero if, for all k ∈ Z, there exist
maps sk: Ak (cid:47)(cid:47)Bk−1 such that for any k we have fk = dk−1 ◦sk +sk+1 ◦dk . Two maps f
B• A•
and g are homotopic if f −g is homotopic to zero.
Proposition 2.4. Being homotopic is an equivalence relation. The elements homotopic to zero
form an “ideal”.
Proof. The first statement is trivial. As for the second: Take, for example, a morphism
f: A• (cid:47)(cid:47)B• and a morphism g: B• (cid:47)(cid:47)C• which is homotopic to zero. We have to check
that g◦f is homotopic to zero. By definition, gk = dk−1◦sk +sk+1◦dk for all k. But then
C• B•
gk ◦fk = dk−1◦sk ◦fk +sk+1◦dk ◦fk = dk−1◦sk ◦fk +sk+1fk+1dk .
C• B• C• C•
Setting (s(cid:48))k = sk◦fk for all k ∈ Z shows that g◦f is homotopic to zero. The proof that g◦f
is homotopic to zero if f is such is completely analogous and left to the reader. (cid:3)
This result allows us to define a new category as follows.
Definition. Let A be an additive category and Kom(A) the category of complexes over A.
The homotopy category K(A) has the same objects as Kom(A) and for any two objects A•
and B• the space of morphisms is Hom (A•,B•) := Hom (A•,B•)/ ∼, where ∼ is the
K(A) Kom(A)
equivalence relation of being homotopic. We denote the appropriate bounded subcategories by
K∗(A), where ∗ ∈ {−,+,b}.
Remark 2.5. The shift functor clearly gives an autoequivalence of K(A).
Definition. Let f: A• (cid:47)(cid:47)B• be a morphism in Kom(A). The mapping cone of f, denoted by
C(f), is the following object of Kom(A):
C(f)k = Ak+1⊕Bk
(cid:18)−dk+1 0 (cid:19)
dk = A•
C(f) fk+1 dk
B•
One can easily check that this is well-defined, that is, C(f) is really a complex.
We have morphisms α(f): B• (cid:47)(cid:47)C(f) and β(f): C(f) (cid:47)(cid:47)A•[1], where α(f) is the natural
injectionandβ(f)isthenaturalprojection. Itisagaineasilycheckedthatthesearemorphisms
of complexes.
Lemma 2.6. For any f: A• (cid:47)(cid:47)B• in Kom(A) there exists a φ: A•[1] (cid:47)(cid:47)C(α(f)) such that
(1) φ is an isomorphism in K(A).
10 P.SOSNA
(2) The following diagram commutes in K(A):
B• α(f)(cid:47)(cid:47) C(f) β(f) (cid:47)(cid:47) A•[1] −f[1] (cid:47)(cid:47) B•[1]
id id φ id
(cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15) (cid:15)(cid:15)
B• α(f)(cid:47)(cid:47) C(f) α(α(f)(cid:47)(cid:47) )C(α(f))β(α(f))(cid:47)(cid:47) B•[1]
Proof. First note that
C(α(f))k = Bk+1⊕C(f)k = Bk+1⊕Ak+1⊕Bk.
Define φk: A[1]k (cid:47)(cid:47)C(α(f))k and ψk: C(α(f))k (cid:47)(cid:47)A[1]k by
−fk+1
φk = idAk+1
0
and
ψk = (0,id ,0).
Ak+1
Then we have the following
(1) φ and ψ are morphisms of complexes.
(2) ψ◦φ = id .
A•[1]
(3) φ◦ψ is homotopic to id .
C(α(f))
(4) ψ◦α(α(f)) = β(f).
(5) β(α(f))◦φ = −f[1].
For (1) note that the differential of C(α(f)) can be written as
−dk+1 0 0 
B•
dk =  0 −dk+1 0 
C(α(f)) A•
idk+1 fk+1 dk
B• B•
Items (2), (4) and (5) are trivial. To get (3) one defines sk: C(α(f))k (cid:47)(cid:47)C(α(f))k−1 by
 
0 0 id
Bk
sk = 0 0 0 
0 0 0
and checks
id −φk ◦ψk(bk+1,ak+1,bk) = (bk+1,ak+1,bk)−(−fk+1(ak+1),ak+1,0)
C(α(f))k
which is equal to
sk+1◦dk (bk+1,ak+1,bk)+dk−1 ◦sk(bk+1,ak+1,bk),
C(α(f)) C(α(f))
since the first term is (bk+1+fk+1(ak+1)+dk(bk),0,0) and the second is (−dkbk,0,bk). (cid:3)